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A386475
Least prime starting a run of exactly n consecutive primes with identical counts of odd digits.
1
2, 5, 3, 11, 97, 503, 499, 491, 14303, 14293, 157259, 157253, 1525723, 4576997, 4576993, 4576991, 10411013, 33388093, 188332121, 194259301, 2853982501, 2853982499, 2853982477, 3913474277, 10883385143, 22809734971, 34883348389, 34883348369, 34883348341
OFFSET
1,1
COMMENTS
Consecutive previous primes of the selected prime are ignored even though the may also have an identical count of digits.
EXAMPLE
a(2) = 5, because the two primes in the sequence starting at 5, namely 5 and 7, each contain the same number of odd digits, and no earlier prime sequence meets this criterion.
In [2], each number contains 0 odd digits.
In [5, 7], each number contains 1 odd digit.
In [3, 5, 7], each number contains 1 odd digit.
In [11, 13, 17, 19], each number contains 2 odd digits.
In [97, 101, 103, 107, 109], each number contains 2 odd digits.
In [503, 509, 521, 523, 541, 547], each number contains 2 odd digits.
In [499, 503, 509, 521, 523, 541, 547], each number contains 2 odd digits.
In [491, 499, 503, 509, 521, 523, 541, 547], each number contains 2 odd digits.
In [14303, 14321, 14323, 14327, 14341, 14347, 14369, 14387, 14389], each number contains 3 odd digits.
In [14293, 14303, 14321, 14323, 14327, 14341, 14347, 14369, 14387, 14389], each number contains 3 odd digits.
MATHEMATICA
oddn[n_] := Plus @@ Mod[IntegerDigits@ n, 2]; T = Table[0, {99}]; p = 1; While[p < 2 10^6, p = NextPrime[p]; c = oddn[p]; r=1; q=p; While[True, q = NextPrime[q]; If[oddn[q] == c, r++, Break[]]]; If[T[[r]] == 0, T[[r]] = p]]; Take[T, Position[T, 0][[1, 1]] - 1] (* Giovanni Resta, Jul 23 2025 *)
CROSSREFS
Sequence in context: A225258 A162613 A120858 * A124937 A279342 A169852
KEYWORD
nonn,base
AUTHOR
Jean-Marc Rebert, Jul 23 2025
EXTENSIONS
More terms from Giovanni Resta, Jul 23 2025
STATUS
approved